loj10136. 「一本通 4.4 练习 3」聚会

思路：

　　声明一下，我这道题并没有AC，只得了九十分，最后一个点会超时，暂时没有找到更好的优化方案。

　　这道题中所求三个点x, y, z的最短路径手推一下就会发现：

　　　　令a = LCA(x, y), b = LCA(y, z), c = LCA(x, z)

　　　　则路径长度为deep[x]+deep[y]+deep[z]-deep[c]-deep[a]-deep[b];

　　　　若a = b， 则聚会点在c, 若b=c，聚会点在a，若a=c，聚会点在b

#include<cstdio>
#include<iostream>
#include<vector>
#include<cmath>
using namespace std;
const int maxn = 500010;
inline void qread(int &x){
    x = 0;
    register int ch = getchar();
    while(ch < '0' || ch > '9')    
        ch = getchar();
    while(ch >= '0' && ch <= '9')
        x = 10 * x + ch - 48, ch = getchar();
}
int n, m, rt = 1;
vector<int> G[maxn];
vector<int> W[maxn];
int llog[maxn];
int f[maxn][25];
int g[maxn];
inline void init(){
    qread(n);    qread(m);
    for(register int i=1; i<n; ++i){
        register int x, y;
        qread(x), qread(y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    g[rt] = 1;
}
void dfs(int x){
    register int sz = G[x].size();
    for(register int i=0; i < sz; ++i)
        if(!g[G[x][i]]){
            f[G[x][i]][0] = x;
            g[G[x][i]] = g[x] + 1;
            dfs(G[x][i]);
        }
}
inline void STtree(){
    for(register int i=2; i<=n; ++i)
        llog[i] = llog[i >> 1] + 1;
    for(register int j = 1; j <= llog[n]; ++j)
        for(register int i=1; i<=n; ++i)
            f[i][j] = f[f[i][j-1]][j-1];        
}
inline int LCA(register int x, register int y){
    if(x == y)     return x;
    if(g[x] < g[y])    swap(x, y);
    while(g[x] > g[y]){
        x = f[x][llog[g[x] - g[y]]];
    }
    if(x == y)     return x;
    for(register int j = llog[n]; j>=0; --j)
        if(f[x][j] != f[y][j]){
            x = f[x][j];
            y = f[y][j];
        }
    return f[x][0];            
}
int main(void){
    init();
    dfs(rt);
    STtree();
    while(m--){
        register int x, y, z;
        qread(x), qread(y), qread(z);
        int a = LCA(x, y), b = LCA(y, z), c = LCA(x, z);
        if(a == b)        printf("%d %d\n", c, g[x] + g[y] + g[z] - g[c] - (g[a] << 1));
        else if(b == c)    printf("%d %d\n", a, g[x] + g[y] + g[z] - g[a] - (g[b] << 1));
        else             printf("%d %d\n", b, g[x] + g[y] + g[z] - g[b] - (g[a] << 1));
    }
}